POJ 2752 Seek the Name, Seek the Fame
来源:互联网 发布:淘宝女装夏装连衣裙 编辑:程序博客网 时间:2024/04/29 05:30
Seek the Name, Seek the Fame
Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 15829 Accepted: 8038
Description
The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names to their newly-born babies. They seek the name, and at the same time seek the fame. In order to escape from such boring job, the innovative little cat works out an easy but fantastic algorithm:
Step1. Connect the father's name and the mother's name, to a new string S.
Step2. Find a proper prefix-suffix string of S (which is not only the prefix, but also the suffix of S).
Example: Father='ala', Mother='la', we have S = 'ala'+'la' = 'alala'. Potential prefix-suffix strings of S are {'a', 'ala', 'alala'}. Given the string S, could you help the little cat to write a program to calculate the length of possible prefix-suffix strings of S? (He might thank you by giving your baby a name:)
Step1. Connect the father's name and the mother's name, to a new string S.
Step2. Find a proper prefix-suffix string of S (which is not only the prefix, but also the suffix of S).
Example: Father='ala', Mother='la', we have S = 'ala'+'la' = 'alala'. Potential prefix-suffix strings of S are {'a', 'ala', 'alala'}. Given the string S, could you help the little cat to write a program to calculate the length of possible prefix-suffix strings of S? (He might thank you by giving your baby a name:)
Input
The input contains a number of test cases. Each test case occupies a single line that contains the string S described above.
Restrictions: Only lowercase letters may appear in the input. 1 <= Length of S <= 400000.
Restrictions: Only lowercase letters may appear in the input. 1 <= Length of S <= 400000.
Output
For each test case, output a single line with integer numbers in increasing order, denoting the possible length of the new baby's name.
Sample Input
ababcababababcababaaaaa
Sample Output
2 4 9 181 2 3 4 5
Source
POJ Monthly--2006.01.22,Zeyuan Zhu
给定一个字符串s,从小到大输出s中既是前缀又是后缀的子串的长度。若s[next[n-1]] == s[n-1],则子串s[0,1,2,...,next[n-1]]是满足条件的子串。然后判断s[next[next[n-1]]] == s[n-1]是否成立,........直到next[next[.....next[n-1]]] == -1为止。
#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define aa 400001int Next[aa];char str[aa];int ans[aa];int cnt;int len;int i,j,t;void getNext(){ Next[0]=-1; j=-1,i=0; while(i<len) { if(j==-1||str[i]==str[j]) { i++; j++; Next[i]=j; } else j=Next[j]; }}int main(){ while (scanf("%s", str) != EOF) { len = strlen(str); getNext(); int ii = 0; t = len; while (t != -1) { ans[ii++] = t ; t = Next[t]; } printf("%d", ans[ii-2]); for (i = ii-3; i >=0; --i) { printf(" %d", ans[i]); } printf("\n"); } return 0;}
0 0
- poj seek the name,seek the fame
- poj---Seek the Name, Seek the Fame
- Seek the Name, Seek the Fame POJ
- Seek the Name, Seek the Fame POJ
- POJ-2752 Seek the Name, Seek the Fame
- poj 2752 Seek the Name, Seek the Fame
- poj 2752 Seek the Name, Seek the Fame(KMP)
- POJ 2752--Seek the Name, Seek the Fame (next)
- POJ:2752Seek the Name, Seek the Fame
- POJ 2752 - Seek the Name, Seek the Fame
- poj 2752 Seek the Name, Seek the Fame
- poj 2752 Seek the Name, Seek the Fame
- POJ-2752-Seek the Name, Seek the Fame
- POJ 2752 Seek the Name, Seek the Fame
- poj 2752 Seek the Name, Seek the Fame
- poj 2752 Seek the Name, Seek the Fame
- POJ 2752 Seek the Name, Seek the Fame
- poj 2752 Seek the Name, Seek the Fame---kmp
- java对象传输流C/S传输对象
- 迅为4412烧写
- 【HDU】 1027 Ignatius and the Princess II
- 润乾集算报表呈现输出之变动行高列宽
- AutoCompleteTextView自动完成文本框简单介绍
- POJ 2752 Seek the Name, Seek the Fame
- 大小转换成M
- Android NDK
- Sqoop中文手册
- CSS overflow显示内容溢出
- 斐波拉契数列不一样的实现
- Android网络编程概述
- ovirt挂载多存储环境时注意事项
- android背景选择器selector用法汇总